Prove each of the following statements below using one of the proof techniques and state the proof strategy you use.

a. The product of an odd and even integer is even.

b. Let m and n be integers. Show that if mn is even, then m is even or n is even.

c. If r is a nonzero rational number and p is an irrational number, then rp is irrational.

d. For all real numbers a, b, and c, max (a, max (b, c)) = max (max (a, b), c).

e. If a and bare rational numbers, then ab is rational too.

f. If a and bare two distinct rational numbers, then there exists an irrational number between them.

g. If m + n and n+p are even integers where m, n, p are integers, then m + p is even.

See below

Step-by-step explanation:

a) Direct proof: Let m be an odd integer and n be an even integer. Then, there exist integers k, j such that m=2k+1 and n=2j. Then mn = (2k+1) (2j) = 2r, where r=j (2k+1) is an integer. Thus, mn is even.

b) Proof by counterpositive: Suppose that m is not even and n is not even. Then m is odd and n is odd, that is, m=2k+1 and n=2j+1 for some integers k, j. Thus, mn=4kj+2k+2j+1=2 (kj+k+j) + 1=2r+1, where r=kj+k+j is an integer. Hence mn is odd, i. e, mn is not even. We have proven the counterpositive.

c) Proof by contradiction: suppose that rp is NOT irrational, then rp=m/n for some integers m, n, n≠. Since r is a non zero rational number, r=a/b for some non-zero integers a, b. Then p=rp/r=rp (b/a) = (m/n) (b/a) = mb/na. Now n, a are non zero integers, thus na is a non zero integer. Additionally, mb is an integer. Therefore p is rational which is contradicts that p is irrational. Hence np is irrational.

d) Proof by cases: We can verify this directly with all the possible orderings for a, b, c. There are six cases:

a≥b≥c, a≥c≥b, b≥a≥c, b≥c≥a, c≥b≥a, c≥a≥b

Writing the details for each one is a bit long. I will give you an example for one case: suppose that c≥b≥a then max (a, max (b, c)) = max (a, c) = c. On the other hand, max (max (a, b), c) = max (b, c) = c, hence the statement is true in this case.

e) Direct proof: write a=m/n and b=p/q, with m, q integers and n, q nonnegative integers. Then ab=mp/nq. mp is an integer, and nq is a non negative integer. Hence ab is rational.

f) Direct proof. By part c), √2/n is irrational for all natural numbers n. Furthermore, a is rational, then a+√2/n is irrational. Take n large enough in such a way that b-a>√2/n (b-a>0 so it is possible). Then a+√2/n is between a and b.

g) Direct proof: write m+n=2k and n+p=2j for some integers k, j. Add these equations to get m+2n+p=2k+2j. Then m+p=2k+2j-2n=2 (k+j-n) = 2s for some integer s=k+j-n. Thus m+p is even.